Question

A horizontal platform with an angular placed on it is executing `S.H.M.` in the vertical direction .The amplitude of oscillation is `4.0 xx 10^(-3)m` What must be the loast period of there oscillation so that the object is not setached from the plateform ? (Taking `g = 10m//s^(2)`)
A. `(pi)/(25) sec`
B. `(pi)/(18) sec`
C. `(pi)/(14) sec`
D. `(pi)/(20) sec`

Answer 1

Correct Answer - A
By drawing free body diagram of object during the exterme position motion at exterme position for equilibrium of mass

`mg - R = mA (A = "Acceleration")`
For critical condition `R = 0`
so `mg = mA rArr mg = ma omega^(2)`
`rArr omega sqrt(g//a) = sqrt((10)/(4.0 xx 10^(-3))) = 50`
`rArr T = (2pi)/(omega) = (2pi)/(50) = (pi)/(25) sec`.